I ran into this graph theory/combinatorics problem, and I'm having some troubles finding the solution.
Considering $K_{10}$, the complete graph with $10$ vertices, what is the number of different matchings in $K_{10}$ with exactly 4 edges? (Two matchings are different if it exists at least one edge that belongs to one matching and not to the other).
If we differ matchings by their sets of vertices then a number of four-edge matchings of $K_n$ is equal to a number of all eight-vertex subsets of $K_{10}$, that is, as ramseysdream111 noticed, ${10\choose 8}={10\choose 2}=45$. But this doesn’t fit the proposed solution. This solution fits for a natural way to differ graphs (in particular, matchings) on a common set of vertices. Namely, such graphs are equal, if their sets of edges coincide. Given a four-edge matching on $K_10$, there are ${10\choose 2}$ possibilities to pick endvertices for its first edge, ${8\choose 2}$ possibilities to pick endvertices for its second edge, ${6\choose 2}$ possibilities to pick endvertices for its third edge, and ${4\choose 2}$ possibilities to pick endvertices for its third edge. So there are in total $N={10\choose 2}{8\choose 2}{6\choose 2}{4\choose 2}$ possibilities to pick an ordered $4$-tuple of edges of a matching, and so $N/4!$ possibilities to pick an unordered $4$-tuple of edges of a matching.